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12 tháng 4 2016

125% . (-1/2)^2 : (1 và 5/6-1,5)+2016^0=5/4 . 1/4 : 11/6 - 3/2 +1

                                                        = 5/16 : 11/6 - 3/2 +1

                                                        =15/88 - 3/2 +1

                                                        = -117/88 + 1

                                                        = -29/88

12 tháng 4 2016

31/16

3 tháng 5 2020

\(125\%\cdot\left(-\frac{1}{2}\right)^2\div\left(1\frac{5}{6}-1,5\right)+2016^0\)

\(=\frac{125}{100}\cdot\frac{1}{4}\div\left(\frac{11}{6}-\frac{3}{2}\right)+1\)

\(=\frac{5}{4}\cdot\frac{1}{4}\div\left(\frac{11}{6}-\frac{9}{6}\right)+1\)

\(=\frac{5}{16}\div\frac{2}{6}+1\)

\(=\frac{15}{16}+1=\frac{31}{16}\)

3 tháng 5 2020

         \(125\%.\left(\frac{-1}{2}\right)^2:\left(1\frac{5}{6}-1,5\right)+2016^0\)

\(=\frac{125}{100}.\frac{1}{4}:\left(\frac{11}{6}-\frac{3}{2}\right)+1\)

\(=\frac{5}{16}:\left(\frac{11}{6}-\frac{9}{6}\right)+1\)

\(=\frac{5}{16} :\frac{1}{3}+1\)

\(=\frac{5}{16} .\frac{3}{1}+1\)

\(=\frac{15}{16}+1\)

\(=\frac{31}{16}\)

\(=\dfrac{5}{4}\cdot\dfrac{1}{4}:\left(\dfrac{21}{16}-\dfrac{24}{16}\right)+1=\dfrac{5}{16}\cdot\dfrac{16}{-3}+1=\dfrac{-5}{3}+1=-\dfrac{2}{3}\)

4 tháng 4 2023

\(125\%\cdot\left(-\dfrac{1}{2}\right)^2:\left(1\dfrac{5}{16}-1,5\right)+2018^0\)
\(=\dfrac{5}{4}\cdot\dfrac{1}{4}:\left(\dfrac{21}{16}-\dfrac{3}{2}\right)+1\)
\(=\dfrac{5}{16}:\left(-\dfrac{3}{16}\right)+1\)
\(=\dfrac{5}{16}\cdot\left(-\dfrac{16}{3}\right)+1\)
\(=-\dfrac{5}{3}+1\)
\(=-\dfrac{5}{3}+\dfrac{3}{3}\)
\(=-\dfrac{2}{3}\)

4 tháng 4 2023

`125%*( (-1)/2 )^2 :(1 5/16 -1,5 )+2018^0`

`=1,25 * 1/4 :(21/16-3/2)+1`

`=1,25*1/4:(21/16-24/16)+1`

`=1,25*1/4:(-3)/16+1`

`=5/16*16/(-3)+1`

`=-5/3+1`

`=-5/3+3/3`

`=-2/3`

8 tháng 6 2019

Bài 1:

\(a,22\frac{1}{2}.\frac{7}{9}+50\%-1,25\)

=\(\frac{45}{2}.\frac{7}{9}+\frac{1}{2}-\frac{5}{4}\)

=\(\frac{35}{2}+\frac{1}{2}-\frac{5}{4}\)

=\(\frac{70}{4}+\frac{2}{4}-\frac{5}{4}\)

=\(\frac{67}{4}\)

\(b,1,4.\frac{15}{49}-\left(\frac{4}{5}+\frac{2}{3}\right):2\frac{1}{5}\)

=\(\frac{7}{5}.\frac{15}{49}-\left(\frac{12}{15}+\frac{10}{15}\right):\frac{11}{5}\)

=\(\frac{3}{7}-\frac{22}{15}.\frac{5}{11}\)

=\(\frac{3}{7}-\frac{2}{3}\)

=\(-\frac{5}{21}\)

\(c,125\%.\left(-\frac{1}{2}\right)^2:\left(1\frac{5}{6}-1,6\right)+2016^0\)

=\(\frac{5}{4}.\frac{1}{4}:\left(\frac{11}{6}-\frac{8}{5}\right)+1\)

=\(\frac{5}{16}:\frac{7}{30}+1\)

=\(\frac{131}{56}\)

\(d,1,4.\frac{15}{49}-\left(20\%+\frac{2}{3}\right):2\frac{1}{5}\)

=\(\frac{7}{5}.\frac{15}{49}-\left(\frac{1}{5}+\frac{2}{3}\right):\frac{11}{5}\)

=\(\frac{3}{7}-\frac{13}{15}:\frac{11}{5}\)

=\(\frac{3}{7}-\frac{13}{33}\)

=\(\frac{8}{231}\)

Bài đ làm giống hệt như bài c

Bài 2 :

\(a,\left|\frac{3}{4}.x-\frac{1}{2}\right|=\frac{1}{4}\)

=>\(\left[{}\begin{matrix}\frac{3}{4}.x-\frac{1}{2}=\frac{1}{4}\\\frac{3}{4}.x-\frac{1}{2}=-\frac{1}{4}\end{matrix}\right.\)

=>\(\left[{}\begin{matrix}\frac{3}{4}.x=\frac{1}{4}+\frac{1}{2}=\frac{3}{4}\\\frac{3}{4}.x=-\frac{1}{4}+\frac{1}{2}=\frac{1}{4}\end{matrix}\right.\)

=>\(\left[{}\begin{matrix}x=\frac{3}{4}:\frac{3}{4}=1\\x=\frac{1}{4}:\frac{3}{4}=\frac{1}{3}\end{matrix}\right.\)

Vậy x ∈{1;\(\frac{1}{3}\)}

\(b,\frac{5}{3}.x-\frac{2}{5}.x=\frac{19}{10}\)

=>\(\frac{19}{15}.x=\frac{19}{10}\)

=>\(x=\frac{19}{10}:\frac{19}{15}=\frac{3}{2}\)

Vậy x ∈ {\(\frac{3}{2}\)}

c,\(\left|2.x-\frac{1}{3}\right|=\frac{2}{9}\)

=>\(\left[{}\begin{matrix}2.x-\frac{1}{3}=\frac{2}{9}\\2.x-\frac{1}{3}=-\frac{2}{9}\end{matrix}\right.\)

=>\(\left[{}\begin{matrix}2.x=\frac{2}{9}+\frac{1}{3}=\frac{5}{9}\\2.x=-\frac{2}{9}+\frac{1}{3}=\frac{1}{9}\end{matrix}\right.\)

=>\(\left[{}\begin{matrix}x=\frac{5}{9}:2=\frac{5}{18}\\x=\frac{1}{9}:2=\frac{1}{18}\end{matrix}\right.\)

Vậy x∈{\(\frac{5}{18};\frac{1}{18}\)}

\(d,x-30\%.x=-1\frac{1}{5}\)

=\(70\%x=-\frac{6}{5}\)

=\(\frac{7}{10}.x=-\frac{6}{5}\)

=>\(x=-\frac{6}{5}:\frac{7}{10}=-\frac{12}{7}\)

Vậy x∈{\(-\frac{12}{7}\)}

8 tháng 6 2019

Bài 2

a/

\(\Rightarrow\left[{}\begin{matrix}\frac{3}{4}.x-\frac{1}{2}=\frac{1}{4}\\\frac{3}{4}.x-\frac{1}{2}=-\frac{1}{4}\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}\frac{3}{4}.x=\frac{1}{4}+\frac{1}{2}\\\frac{3}{4}.x=-\frac{1}{4}+\frac{1}{2}\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}\frac{3}{4}.x=\frac{3}{4}\\\frac{3}{4}.x=\frac{1}{4}\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x=\frac{3}{4}:\frac{3}{4}\\x=\frac{1}{4}:\frac{3}{4}\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x=1\\x=\frac{1}{3}\end{matrix}\right.\)

Vậy \(x=1\) hoặc \(x=\frac{1}{3}\)

b/ Đặt x làm thừa số chung rồi tính như bình thường

c/ Tương tự câu a

d/ Tương tự câu b